Point of tangency between given linear equation and circle equation

Question

I'm trying to find the point of tangency between the circle $ (x-3)^2 + (y-4)^2=25 $ and the linear equation $ y = -0.75x + 3.125 $ as seen on the graph.

In the image I attached I used trial and error to get the approximate point of tangency between the linear curve and the circle which is almost at $ 18.75 = 25-(x-3)^2 - (y-4)^2 $

Does anyone know how to solve this? I'm trying to find the tangency point a circle shaped utility function and a a budget restriction (linear equation) for a micro economics class.

Answer 1

The direction of the gradient vector at the point of tangency should be the same as the direction of normal to the circle.

The gradient vector is $$ <2(x-3),2(y-4)>$$ and the slope of the normal vector is the opposite reciprocal of the slope of the tangent line.

That is $$\frac {y-4}{x-3} = 4/3$$ The above equation implies $$y=(4/3)x$$ Plug in the equation of the given line, $$ y=-.75x+3.125$$ We get the point of tangency to be $$(x,y)= (1.5,2)$$ as indicated by the graph.

Answer 2

We have

$$ \cases{ (x-x_0)^2+(y-y_0)^2 = r^2\\ a x + b y = c } $$

after substitution we have

$$ (x-x_0)^2+((c-a x)/b -y_0)^2= r^2 $$

and then the solution is at

$$ x = \frac{b \left(b x_0\pm \sqrt{r^2 \left(a^2+b^2\right)-(a x_0+b y_9-c)^2}\right)+a (c-b y_0)}{a^2+b^2} $$

now, at tangency $x$ should be unique thus

$$ r^2 \left(a^2+b^2\right)-(a x_0+b y_9-c)^2 = 0 $$

and solving for $c$ it gives

$$ c = \pm r \sqrt{\left(a^2+b^2\right)}+a x_0+b y_0 $$